Question

Question #eddee

Answer 1

`d^n/(dx)^n(1/x) = (-1)^n(n!)/x^(n+1)`

Explanation:

We have that:

`(1) f^((n))(x) = (-1)^n(n!)x^(-(n+1))`

We can prove it directly for `n=1`:

`f'(x) = d/dx(1/x) = -1/x^2 = (-1)^1(1!)x^(-(1+1))`

In general, suppose:

`f^((n))(x) = (-1)^n(n!)x^(-(n+1))`

we have that:

`f^((n+1))(x) = d/dx f^((n))(x) = (-1)^n(n!)(-(n+1))x^(-(n+2)) = (-1)^(n+1)((n+1)!)x^(-(n+2))`

So the equation (1) is proven by induction.